Correct. Rewrite in standard form by subtracting $\displaystyle{4}{x}+{15}\usepackage{cmbright}\Huge\$ from both sides: $\displaystyle{{x}}^{{2}}+{2}{x}-{15}={0}\usepackage{cmbright}\Huge\$

Use the Quadratic Formula, $\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\usepackage{cmbright}\Huge\$ , with $\displaystyle{a}={1},{b}={2}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}=-{15}\usepackage{cmbright}\Huge\$ :

$\displaystyle{x}=\frac{{-{2}\pm\sqrt{{{{2}}^{{2}}-{4}{\left({1}\right)}{\left(-{15}\right)}}}}}{{{2}\cdot{1}}}=\frac{{-{2}\pm\sqrt{{{4}+{60}}}}}{{2}}=\frac{{-{2}\pm\sqrt{{64}}}}{{{2}}}=\frac{{-{2}\pm{8}}}{{{2}}}\usepackage{cmbright}\Huge\$

So $\displaystyle{x}=\frac{{-{2}+{8}}}{{2}}=\frac{{6}}{{2}}={3}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=\frac{{-{2}-{8}}}{{2}}=\frac{{-{10}}}{{2}}=-{5}\usepackage{cmbright}\Huge\$